Question

Analysis of a volatile liquid shows that it contains 23.81 percent carbon, 5.99 percent hydrogen, and 70.20 percent chlorine by mass. At 260.0 °C and 4.7 atm, 186.0 mL of the vapor has a mass of 1.009 g. What is the molar mass of the compound?

Answer 1

A. Molar mass by ideal gas law

Using formula,

moles (n) = PV/RT

with,

P = 4.7 atm

V = 0.186 L

R = 0.08205 L.atm/mol.K

T = 260 oC = 260 + 273 = 533 K

Feed values,

n = 4.7 x 0.186/0.08205 x 533

   = 0.020 mols

moles = g/molar mass

molar mass = 1.009 g/0.020 mol = 50.45 g/mol

B. Molar mass by using the other data we can find empirical formula,

moles of components,

C = 0.2381/12 = 0.0198

H = 0.0599/1 = 0.0599

Cl = 0.7020/35.5 = 0.0198

divide by smallest factor,

C = 0.0198/0.0198 = 1

H = 0.0599/0.0198 = 3

Cl = 0.0198/0.0198 = 1

So empirical formula = CH3Cl

empirical formula mass = 12 + 3 x 1 + 35.5 = 50.5 g/mol

So molar mass by both methods is 50.5 g/mol